If the angular speed of earth is increased so much that the objects start flying from the equator, then the length of the day will be nearly

The correct option is 1.5 hours.

To understand why this is the correct answer, let us derive the length of the day step-by-step using the principles of rotational mechanics and gravitation.

Step 1: Understanding the effective gravity at the equator
The effective acceleration due to gravity $g_e$ at the equator of the Earth is affected by the Earth's rotation. It is given by the formula:
$g_e=g-{\omega }^{2}R$
where:
- $g$ is the acceleration due to gravity on the surface of the Earth without rotation (approximately $9.8{\mathrm{m/s}}^2$).
- $\omega$ is the angular speed of the Earth's rotation.
- $R$ is the radius of the Earth (approximately $6400\mathrm{km}=6.4\times {10}^{6}m$).

Step 2: Condition for objects to start flying off
For objects to start flying off from the equator, they must experience weightlessness. This means the effective acceleration due to gravity at the equator must become zero:
$g_e=0$
Substituting this condition into our formula:
$0=g-{\omega }^{2}R$
${\omega }^{2}R=g$
$\omega =\sqrt{\frac{g}{R}}$

Step 3: Calculating the new length of the day
The length of the day, which represents the time period $T$ of one full rotation of the Earth, is related to the angular speed $\omega$ by the relation:
$T=\frac{2\pi }{\omega }$
Substituting the expression for $\omega$:
$T=2\pi \sqrt{\frac{R}{g}}$

Now, let's plug in the standard values:
- $R=6.4\times {10}^{6}m$
- $g=9.8{\mathrm{m/s}}^2$
$T=2\pi \sqrt{\frac{6.4\times {10}^6}{9.8}}$
$T\approx 2\times 3.14\times \sqrt{653061.2}$
$T\approx 6.28\times 808.1\mathrm{seconds}$
$T\approx 5075\mathrm{seconds}$

Step 4: Converting seconds into hours
To convert the time period from seconds to hours, we divide by 3600 (since 1 hour = 3600 seconds):
$T=\frac{5075}{3600}\mathrm{hours}$
$T\approx 1.41\mathrm{hours}$
This is approximately equal to 1.5 hours (or more precisely, about 84 minutes).